Positive charges produce an outward-pointing electric field. Thus the given charge must have a positive sign. The magnitude of the electric field due to this charge (with magnitude $|q|$) at a distance $r;(= 0.44;mathrm{m})$, is given by $ُُُE = k|q|/r^2$. Solving for $|q|$ and substituting the known values, we find

$$
begin{align*}
|q| &= frac{Er^2}{k}\
&= frac{(2.7times10^4;mathrm{N/C})(0.44;mathrm{m})^2}{8.99times10^9;mathrm{N}cdotmathrm{m}^2/mathrm{C}^2}\
&= 0.58;mumathrm{C}
end{align*}
$$

We see the charge is $q = +0.58;mu$C.

$q = +0.58;mu$C

Applying $E = k|q|/r^2$, with $E = 0.78times10^4$ N/C and $|q| = 3.1;mu$C, we should solve for $r$ as follows

$$
begin{align*}
r &= sqrt{frac{k|q|}{E}}\
&= sqrt{frac{(8.99times10^9;mathrm{N}cdotmathrm{C}^2/mathrm{N}^2)(3.1times10^{-6};mathrm{C})}{0.78times10^4;mathrm{N/C}}}\
&= 1.9;mathrm{m}
end{align*}
$$